The angle between the curves y = sin x and y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The curves $y = \sin x$ and $y = \cos x$ intersect where $\sin x = \cos x$,which implies $	an x = 1$. Since $0 < x < \frac{\pi}{2}$,the intersect

Vedclass · Accepted Answer

$	an^{-1} (2 \sqrt{2})$

Vedclass · Answer

(C) The curves $y = \sin x$ and $y = \cos x$ intersect where $\sin x = \cos x$,which implies $	an x = 1$. Since $0 Let $m_1$ and $m_2$ be the slopes of the tangents to the curves at $x = \frac{\pi}{4}$.For $y = \sin x$,$\frac{dy}{dx} = \cos x$. Thus,$m_1 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.For $y = \cos x$,$\frac{dy}{dx} = -\sin x$. Thus,$m_2 = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.Substituting the values: $	an 	heta = |\frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})}| = |\frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}}| = |\frac{\sqrt{2}}{\frac{1}{2}}| = 2\sqrt{2}$.Therefore,$	heta = 	an^{-1}(2\sqrt{2})$.

The angle between the curves $y = \sin x$ and $y = \cos x$,$0 < x < \frac{\pi}{2}$,is

Similar Questions

The value of $c$ such that the straight line joining the points $(0,3)$ and $(5,-2)$ is tangent to the curve $y=\frac{c}{x+1}$ is

The sum of the squares of the intercepts on the axes made by a tangent at any point on the curve $x^{2/3} + y^{2/3} = a^{2/3}$ is -

The normal to the curve $x = 9(1 + \cos \theta)$,$y = 9 \sin \theta$ at $\theta$ always passes through the fixed point

The number of points on the curve $y=2t^2+3t-5$ and $x=t^3-4t^2-3t$ such that the normals drawn at them on the curve are parallel to the $X$-axis is

If the normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\left(\frac{3 \pi}{4}\right)^{C}$ with the positive $X$-axis,then $f^{\prime}(3)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module